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Math Help - find the principal unit normal vector to the curve

  1. #1
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    find the principal unit normal vector to the curve

    find the principal unit normal vector to the curve at the specific value of the parameter.

    r(t) = ti + \frac {1}{2}t^2j,  t = 2

    i need to use this formula

     T(t) =  \frac {r'(t)}{ || r'(t) ||}

    so, to find r'(t)


      r'(t) = i + t j

    so, now to find the magnitude: simplified

    || r'(t) || = \sqrt  {1 + t^2}


    puting them together

     T(t) =  \frac {i + t j} {\sqrt {1 + t^2}}

    From here is where i get lost

    I need to find T'(t)

    I can seem to follow the solutions manual. I think they are using the quiotient rule due to the denominator having the 3/2
    fraction. But if they are i still cant see how they are doing it.
    Can anyone help me work out the next step for T'(t)?



    T'(t) =  \frac { -t}{(t^2 + 1)^{3/2}}i + \frac {1}{ (t^2 + 1) ^{3/2}} j

    Thank you so much!!!
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Re: find the principal unit normal vector to the curve

    \mathbf{T}(t) = \frac{\mathbf{i} + t\mathbf{j}}{\sqrt{1 + t^2}}

    \Rightarrow\mathbf{T}(t) = \left(1+t^2\right)^{-1/2}\mathbf{i} + t\left(1+t^2\right)^{-1/2}\mathbf{j}

    Differentiating,

    \mathbf{T}'(t) = -\frac12(2t)\left(1+t^2\right)^{-3/2}\mathbf{i} + \left[-\frac12t(2t)\left(1+t^2\right)^{-3/2} + \left(1+t^2\right)^{-1/2}\right]\mathbf{j}

    \Rightarrow\mathbf{T}'(t) = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \left[\frac{-t^2}{\left(1+t^2\right)^{3/2}} + \frac1{\left(1+t^2\right)^{1/2}}\right]\mathbf{j}

    \Rightarrow\mathbf{T}'(t) = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac{-t^2+\left(1+t^2\right)}{\left(1+t^2\right)^{3/2}}\mathbf{j}

    \Rightarrow\mathbf{T}'(t) = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac1{\left(1+t^2\right)^{3/2}}\mathbf{j}

    We find the magnitude,

    \left\|\mathbf{T}'(t)\right\| = \sqrt{\frac{t^2+1}{\left(1+t^2\right)^3}

    \Rightarrow\left\|\mathbf{T}'(t)\right\| = \sqrt{\frac1{\left(t^2+1\right)^2}

    \Rightarrow\left\|\mathbf{T}'(t)\right\| = \frac1{t^2+1}

    And therefore,

    \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\left\|\mathbf{T}'(t)\right \|}

    =\left(t^2+1\right)\left[\frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac1{\left(1+t^2\right)^{3/2}}\mathbf{j}\right]

    =\frac{-t}{\sqrt{1+t^2}}\mathbf{i} + \frac1{\sqrt{1+t^2}}\mathbf{j}\right]
    Last edited by Reckoner; June 14th 2012 at 07:39 PM.
    Thanks from icelated
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    Re: find the principal unit normal vector to the curve

    in your second step the first arrow before Differentiating, im not sure how you got j

    for i did you just bring up the denominator? doesnt make sense
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    MHF Contributor Reckoner's Avatar
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    Re: find the principal unit normal vector to the curve

    Quote Originally Posted by icelated View Post
    in your second step the first arrow before Differentiating, im not sure how you got j

    for i did you just bring up the denominator? doesnt make sense
    Did you forget these properties of exponents from algebra?

    x^{-n} = \frac1{x^n}

    x^{m/n} = \sqrt[n]{x^m}
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  5. #5
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    Re: find the principal unit normal vector to the curve

    Wow, okay how did i miss that! I wasnt splitting the i and j up. Once i did now i can see the answer fall out.
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